Evaluate the definite integral. $\int^1_{-1}\big(9x^2+6x-3\big)\,dx = $
Solution: First, use the power rule: $\int^1_{-1}\big(9x^2+6x-3\big)\,dx~=~3x^3+3x^2-3x\Bigg|^1_{{-1}}$ Second, plug in the limits of integration: $[3\cdot1^3+3\cdot1^2-3\cdot1]-[3\cdot({-1})^3+3\cdot({-1})^2-3\cdot({-1})] = 3-3 = 0$. The answer: $\int^1_{-1}\big(9x^2+6x-3\big)\,dx = 0$